initial commit / add all books

4 files changed, 307 insertions, 0 deletions

diff --git a/905/CH5/EX5.1/5_1.sce b/905/CH5/EX5.1/5_1.sce
new file mode 100755
index 000000000..b9f08f375
--- /dev/null
+++ b/905/CH5/EX5.1/5_1.sce
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+clear;

+clc;

+

+// Illustration 5.1

+// Page: 287

+

+printf('Illustration 5.1 -  Page: 287\n\n');

+

+// solution

+

+//*****Data*****//

+// Component 'A' is to be absorbed //

+y_N1 = 0.018; // [mole fraction 'A' of in entering gas]

+y_1 = 0.001; // [mole fractio of 'A'in leaving gas]

+x_0 = 0.0001; // [mole fraction of 'A' in entering liquid]

+m = 1.41; // [m = yi/xi]

+n_1 = 2.115; // [molar liquid to gas ratio at bottom, L/V]

+n_2 = 2.326; // [molar liquid to gas ratio at top, L/V]

+E_MGE = 0.65;

+//*****//

+

+printf('Illustration 5.1 (a) -  Page: 287\n\n');

+// Solution (a)

+

+A_1 = n_1/m; // [absorption factor at bottom]

+A_2 = n_2/m; // [absorption factor at top]

+

+A = sqrt(A_1*A_2);

+// Using equation 5.3 to calculate number of ideal stages

+N = (log(((y_N1-m*x_0)/(y_1-m*x_0))*(1-1/A) + 1/A))/log(A); // [number of ideal stages]

+printf("Number of ideal trays is %f\n",N);

+// Using equation 5.5

+E_o = log(1+E_MGE*(1/A-1))/log(1/A);

+// Therefore number of real trays will be

+n = N/E_o;

+printf("Number of real trays is %f\n",n);

+n = 8;

+printf("Since it is not possible to specify a fractional number of trays, therefore number of real trays is %f\n\n",n);

+

+printf('Illustration 5.1 (b) -  Page: 287\n\n');

+

+// Solution (b)

+

+// Back checking the answer

+printf('Back checking the answer'); 

+N_o = E_o*n;

+// Putting N_o in equation 5.3 to calculate y_1

+deff('[y] = f16(Z)','y=N_o-(log(((y_N1-m*x_0)/(Z-m*x_0))*(1-1/A) + 1/A))/log(A)');

+Z = fsolve(0.001,f16);

+printf("Mole fraction of A in leaving gas is %f percent which satisfies the requirement that the gas exit concentration should not exceed 0.1 percent.",Z);

+

+// For a tower diameter of 1.5 m, Table 4.3 recommends a plate spacing of 0.6 m

+Z = n*0.6; // [Tower height, m]

+printf("The tower height will be %f m",Z);
\ No newline at end of file
diff --git a/905/CH5/EX5.3/5_3.sce b/905/CH5/EX5.3/5_3.sce
new file mode 100755
index 000000000..9456a43fc
--- /dev/null
+++ b/905/CH5/EX5.3/5_3.sce
@@ -0,0 +1,102 @@
+clear

+clc;

+

+// Illustration 5.3

+// Page: 295

+

+printf('Illustration 5.3 -  Page: 295\n\n');

+

+// solution

+// For tower diameter, packed tower design program of Appendix D is run using // the data from Example 5.2 and packing parameters from Chapter 4.

+

+// For a pressure drop of 300 Pa/m, the program converges to a tower diameter 

+Db = 0.641; // [m]

+// Results at the bottom of tower

+fb= 0.733; // [flooding]

+ahb = 73.52; // [m^-1]

+Gmyb = 126; // [mol/square m.s]

+kyb = 3.417; // [mol/square m.s]

+klb = 9.74*10^-5; // [m/s]

+

+// From equation 2.6 and 2.11

+// Fg = ky*(1-y),   Fl = kx*(1-x)

+// Assume 1-y = 1-y1    1-x = 1-x1

+// let t = 1-y1  u = 1-x1

+// Therefore

+t = 0.926;

+u = 0.676;

+Fgb = kyb*t; // [mol/square m.s]

+rowlb = 780; // [kg/cubic m]

+Mlb = 159.12; // [gram/mole]

+c = rowlb/Mlb; // [kmle/cubic m]

+Flb = klb*c*u; // [mol/square m.s]

+// From equ 5.19

+Htgb = Gmyb/(Fgb*ahb); // [m]

+

+// Now, we consider the conditions at the top of the absorber

+// For a pressure drop of 228 Pa/m, the program converges to a tower        // diameter

+Dt = 0.641; // [m]

+// Results at the top of tower

+ft = 0.668; // [flooding]

+aht = 63.31; // [m^-1]

+Gmyt = 118; // [mol/square m.s]

+kyt = 3.204; // [mol/square m.s]

+klt = 8.72*10^-5; // [m/s]

+

+rowlt = 765; // [kg/cubic m]

+Mlt = 192.7; // [gram/mole]

+cl = rowlt/Mlt; // [kmole/cubic m]

+Fgt = kyt*0.99; // [mole/square m.s]

+Flt = klb*cl*0.953; // [mole/square m.s]

+// From equ 5.19

+Htgt = Gmyt/(Fgt*aht); // [m]

+Htg_avg = (Htgb+Htgt)/2; // [m]

+Fg_avg = (Fgt+Fgb)/2; // [mole/square m.s]

+Fl_avg = (Flb+Flt)*1000/2; // [mole/square m.s]

+

+// The operating curve equation for this system in terms of mole fractions

+// y = 

+

+// From Mathcad program figure 5.3

+x1 = 0.324;

+x2 = 0.0476;

+n = 50;

+dx = (x1-x2)/n;

+me = 0.136;

+T = zeros(50,2);

+for j=1:50

+    x(j) = x2+j*dx;

+    y(j) = (0.004+0.154*x(j))/(1.004-0.846*x(j));

+    

+    deff('[y] = f12(yint)','y = (1-yint)/(1-y(j)) - ((1-x(j))/(1-yint/me))^(Fl_avg/Fg_avg)');

+    yint(j) = fsolve(0.03,f12);

+    f(j) = 1/(y(j)-yint(j));

+    T(j,1) = y(j);

+    T(j,2) = f(j);

+end

+

+scf(1);

+plot(T(:,1),T(:,2));

+xgrid();

+xlabel("y");

+ylabel("f = 1/(y-yint)");

+

+yo = y(1);

+yn = y(50);

+// From graph between f vs y

+Ntg = 10.612;

+// Therefore

+Z = Htg_avg*Ntg; // [m]

+printf("The total packed height is %f m.\n\n",Z);

+deltaPg = 300*Z; // [Pa]

+Em = 0.60; // [mechanical efficiency]

+Qg = 1.0;

+Wg = (Qg*deltaPg)/Em; // [Power required to force the gas through the tower, W]

+L2 = 1.214; // [kg/s]

+g = 9.8; // [m/square s]

+Wl = L2*g*Z/Em; // [Power required to pump the liquid to the top of the absorber, W]

+printf("The power required to force the gas through the tower is %f W.\n\n",Wg);

+printf("The power required to pump the liquid to the top of the absorber is %f W.\n\n",Wl);

+

+

+

diff --git a/905/CH5/EX5.4/5_4.sce b/905/CH5/EX5.4/5_4.sce
new file mode 100755
index 000000000..d3e9c396d
--- /dev/null
+++ b/905/CH5/EX5.4/5_4.sce
@@ -0,0 +1,62 @@
+clear

+clc;

+

+// Illustration 5.4

+// Page: 299

+

+printf('Illustration 5.4 -  Page: 299\n\n');

+

+// solution

+// Fro example 4.4

+m = 0.57;

+D = 0.738; // [tower diameter, m]

+G = 180; // [rate of gas entering the tower, kmole/h]

+L = 151.5; // [rate of liquid leaving the tower, kmole/h]

+// Amount of ethanol absorbed 

+M = G*0.02*0.97; // [kmole/h]

+//*****//

+

+// Inlet gas molar velocity

+Gmy1 = G*4/(3600*%pi*D^2); // [kmole/square m.s]

+// Outlet gas velocity

+Gmy2 = (G-M)*4/(3600*%pi*D^2); // [kmole/square m.s]

+// Average molar gas velocity

+Gmy = (Gmy1+Gmy2)/2; // [kmole/square m.s]

+

+// Inlet liquid molar velocity

+Gmx2 = L*4/(3600*%pi*D^2); // [kmole/square m.s]

+// Outlet liquid molar velocity

+Gmx1 = (L+M)*4/(3600*%pi*D^2); // [kmole/square m.s]

+

+// Absorption factor at both ends of the column:

+A1 = Gmx1/(m*Gmy1);

+A2 = Gmx2/(m*Gmy2);

+// Geometric average

+A = sqrt(A1*A2);

+

+y1 = 0.02;

+// For 97% removal of the ethanol

+y2 = 0.03*0.02;

+// Since pure water is used 

+x2 = 0;

+// From equation 5.24

+Ntog = log((y1-m*x2)/(y2-m*x2)*(1-1/A)+1/A)/(1-1/A);

+

+// From example 4.4

+// ky*ah = 0.191 kmole/cubic m.s

+// kl*ah = 0.00733 s^-1

+kyah = 0.191; // [kmole/cubic m.s]

+klah = 0.00733; // [s^-1]

+rowl = 986; // [kg/cubic m]

+Ml = 18; // [gram/mole]

+c = rowl/Ml; // [kmole/cubic m]

+kxah = klah*c; // [kmole/cubic m.s]

+

+// Overall volumetric mass transfer coefficient

+Kyah = (kyah^-1 + m/kxah)^-1; // [kmole/cubic m.s]

+

+// From equation 5.22

+Htog = Gmy/Kyah; // [m]

+// The packed height is given by equation 5.21,

+Z = Htog*Ntog; // [m]

+printf("The packed height of an ethanol absorber is %f m.\n\n",Z);
\ No newline at end of file
diff --git a/905/CH5/EX5.5/5_5.sce b/905/CH5/EX5.5/5_5.sce
new file mode 100755
index 000000000..7d569dcb9
--- /dev/null
+++ b/905/CH5/EX5.5/5_5.sce
@@ -0,0 +1,89 @@
+clear

+clc;

+

+// Illustration 5.5

+// Page: 302

+

+printf('Illustration 5.5 -  Page: 302\n\n');

+

+// solution

+

+//*****Data*****//

+;// a = CH4 b = C5H12

+Tempg = 27;// [OC]

+Tempo = 0;// [base temp,OC]

+Templ = 35;// [OC]

+xa = 0.75;// [mole fraction of CH4 in gas]

+xb = 0.25;// [mole fraction of C5H12 in gas]

+M_Paraffin = 200;// [kg/kmol]

+hb = 1.884;// [kJ/kg K]

+//********//

+

+Ha = 35.59;// [kJ/kmol K]

+Hbv = 119.75;// [kJ/kmol K]

+Hbl = 117.53;// [kJ/kmol K]

+Lb = 27820;// [kJ/kmol]

+// M = [Temp (OC) m]

+M = [20 0.575;25 0.69;30 0.81;35 0.95;40 1.10;43 1.25];

+// Basis: Unit time

+GNpPlus1 = 1;// [kmol]

+yNpPlus1 = 0.25;// [kmol]

+HgNpPlus1 = ((1-yNpPlus1)*Ha*(Tempg-Tempo))+(yNpPlus1*(Hbv*(Tempg-Tempo)+Lb));// [kJ/kmol]

+L0 = 2;// [kmol]

+x0 = 0;// [kmol]

+HL0 = ((1-x0)*hb*M_Paraffin*(Templ-Tempo))+(x0*hb*(Templ-Tempo));// [kJ/kmol]

+C5H12_absorbed = 0.98*xb;// [kmol]

+C5H12_remained = xb-C5H12_absorbed;

+G1 = xa+C5H12_remained;// [kmol]

+y1 = C5H12_remained/G1;// [kmol]

+LNp = L0+C5H12_absorbed;// [kmol]

+xNp = C5H12_absorbed/LNp;// [kmol]

+// Assume:

+Temp1 = 35.6;// [OC]

+Hg1 = ((1-y1)*Ha*(Temp1-Tempo))+(y1*(Hbv*(Temp1-Tempo)+Lb));// [kJ/kmol]

+

+

+Qt = 0;

+deff('[y] = f30(HlNp)','y = ((L0*HL0)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(G1*Hg1)+Qt)');

+HlNp = fsolve(2,f30);

+

+deff('[y] = f31(TempNp)','y = HlNp-(((1-x0)*hb*M_Paraffin*(TempNp-Tempo))+(x0*hb*(TempNp-Tempo)))');

+TempNp = fsolve(35.6,f31);

+// At Temp = TempNp:

+mNp = 1.21;

+yNp = mNp*xNp;// [kmol]

+GNp = G1/(1-yNp);// [kmol]

+HgNp = ((1-yNp)*Ha*(TempNp-Tempo))+(yNp*(Hbv*(TempNp-Tempo)+Lb));// [kJ/kmol]

+// From equation 5.28 with n = Np-1

+deff('[y] = f32(LNpMinus1)','y = LNpMinus1+GNpPlus1-(LNp+GNp)');

+LNpMinus1 = fsolve(2,f32);// [kmol]

+

+// From equation 5.29 with n = Np-1

+deff('[y] = f33(xNpMinus1)','y = ((LNpMinus1*xNpMinus1)+(GNpPlus1*yNpPlus1))-((LNp*xNp)+(GNp*yNp))');

+xNpMinus1 = fsolve(0,f33);// [kmol]

+

+// From equation 5.30 with n = Np-1

+deff('[y] = f34(HlNpMinus1)','y = ((LNpMinus1*HlNpMinus1)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(GNp*HgNp))');

+HlNpMinus1 = fsolve(0,f34);// [kJ/kmol]

+deff('[y] = f35(TempNpMinus1)','y = HlNpMinus1-(((1-xNpMinus1)*hb*M_Paraffin*(TempNpMinus1-Tempo))+(xNpMinus1*hb*(TempNpMinus1-Tempo)))');

+TempNpMinus1 = fsolve(42,f35);// [OC]

+

+// The computation are continued upward through the tower in this manner until the gas composition falls atleast to 0.00662.

+// Results = [Tray No.(n) Tn(OC) xn yn]

+Results = [4.0 42.3 0.1091 0.1320;3 39.0 0.0521 0.0568;2 36.8 0.0184 0.01875;1 35.5 0.00463 0.00450];

+scf(8);

+plot(Results(:,1),Results(:,4));

+xgrid();

+xlabel('Tray Number');

+ylabel('mole fraction of C5H12 in gas');

+

+scf(9);

+plot(Results(:,1),Results(:,2));

+xgrid();

+xlabel('Tray Number');

+ylabel('Temparature(OC)');

+

+// For the cquired y1

+Np = 3.75;

+printf("The No. of trays will be %f",Np);

+ 
\ No newline at end of file